MATHEMATICS N5

MATHEMATICS N5
M.J.J. van Rensburg
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TROUPANT
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Publishers
Copyright © 2004 by the author
All rights reserved. No part of this publication may be
reproduced or transmitted in any form or by any means
without prior written permission by the publisher.
ISBN: 978 1 919780 84 9; eISBN: 978 1 430802 66 2
First edition 2004
Second impression 2005
Third impression 2006
Fourth impression 2007
Fifth impression 2011
Sixth impression 2011
Seventh impression 2012
Eighth impression 2012
Published by
Troupant Publishers
Suite 10, Private Bag X12
Cresta, 2118
Set in 10 on 12pt Times New Roman
Typesetting by Compleat Typesetters
Printed and bound by Ultra Litho (Pty) Limited
Preface
There is no generally accepted theory of how to teach or learn; there
are only guidelines which often overlap and supplement each other. It
is generally accepted that when you understand something, it is much
easier to learn. It is also true that if you do not initially understand
something, constant repetition will eventually result in understanding.
The emphasis in this book is on understanding. When you understand
something, however, it does not necessarily mean that you will be able
to do it. Ample provision has therefore been made for practice.
One of the characteristics of Mathematics is that it is compact and that
mathematical problems lend themselves to very short solutions.
However, for students to master the subject in a meaningful way,
the material must be covered in greater detail. The first examples in
this book will therefore be set out properly, and will not necessarily
demonstrate the shortest methods of solving problems. It is, after all, a
fact of life that before you can run, you must first learn to walk.
Properly set out examples are also beneficial for exam revision, since
by this time students have often forgotten the short-cut methods and
are unable to work them out on their own.
THE AUTHOR
Contents
Module 1: Limits and continuity
1.1
1.2
1.2.1
1.2.2
1.3
1.4
Theorems on limits
~; 00
Limits of the indeterminate forms
Factorising or dividing limits of the form
Dividing limits of the form ~
L'Hospital's rule
Continuity
g;
g
1
oo;Oxoo 3
3
3
4
7
Module 2: Differentiation
2.1
2.1.1
2.1.2
2.2.
2.2.1
2.2.2
2.2.3
2.2.4
2.2.5
2.2.6
2.2.7
2.2.8
2.2.9
Differentiation from first principles
Algebraic functions
sin x and cos x
Differentiation techniques
Standard differential coefficients (and rules)
The chain rule
Standard differential coefficients for differentiating
functions of functions
The product rule and the quotient rule
Implicit differentiation
Logarithmic differentiation
Differentiating inverse trigonometric functions
(arc functions)
Sketch graphs of the arc functions
Successive differentiation
11
11
11
13
15
15
18
23
28
33
37
39
45
50
Module 3: Applications of differentiation
52
3.1
3.2
52
57
64
3.3
Newton's rule
Maxima and minima
Rate of change and related ratios
Module 4: Integration techniques
4.1
4.1.2
4.1.3
4.1.4
4.2
4.2.1
4.2.2
4.2.3
4.2.4
4.3
Basic integration and algebraic substitutions
Algebraic substitutions
Fractions where the degree of the numerator is higher
than or equal to the degree of the denominator
Integration as the inverse of differentiation
Trigonometric functions
sin 2 ax; cos 2 ax; tan 2 ax and cot 2 ax
sin ax and cos bx
sinm x . cos n x where m and n are uneven and ~ 5
sin and tan substitutions
Integration by parts
71
71
80
83
84
86
86
88
89
90
92
Module 5: Partial fractions
96
5.1
5.2
5.3
96
97
Introduction
Different factors in the denominator
Recurring factors
98
Module 6: Applications of definite integrals
101
6.1
6.1.1
6.1.2
6.1.3
6.1.4
6.1.5
6.1.6
101
101
103
103
103
105
106
Solving definite integrals
Introduction
Interchanging limits
Another property of definite integrals
Changing limits to correspond with change in variable
Infinite limits
Laplace transforms
Module 7: Areas and volumes
110
7.1
7.1.1
7.1.2
7.2
110
110
117
121
Areas
The area bounded by a curve and a reference axis
The area bounded by two curves
Volumes
Module 8: The second moment of area
129
8.1
8.2
8.2.1
8.2.2
8.2.3
129
Introduction
Laminas
The second moment of area of a rectangular lamina
The parallel axis theorem for laminas
The second moment of area of circular laminas
141
Module 9: The moment of inertia
9.1
The relationship between the moment of inertia
and energy
10.1
Introduction
Variables separable
cP
Equations of the form ~ = ax 2
141
151
Module 10: Differential equations
10.2
10.3
130
130
135
137
+ bx + c
151
152
156
Criteria tests
160
Answers to criteria tests
164
List of formulae
168
Module 1
Limits and continuity
Note: If your N4 differentiation is not up to standard it is advisable to do
sections 2.2.1 to 2.2.4 before you do Module 1, because you will
need differentiation techniques to apply L'Hospital's rule.
Objectives and overview
On completion of this module, you should be able to:
1. apply L'Hospital's rule if the quotient is in one of the following
forms'. Q.
00. 00 00'
0' 00'
,
2. express the conditions for continuity; and
3. determine the continuity or discontinuity of a given function at
a given point.
1.1 THEOREMS ON LIMITS
The following theorems were discussed in N4:
1. Lim k(f)x = k Lim fix)
x-a
x-a
2. Lim [f(x) ± g(x)]
= Limf(x) ±
x~a
Lim g(x)
x~a
3. Lim k = k
x~a
4. Lim [ttx) . g(x)]
= Limf(x) . Lim g(x)
x~a
x~a
fix)
5. Lim - ()
x~a
g
X
=
Limf(x)
~
Lim g(x)
where Lim g(x) =I- 0
6. Lim ytf(x) = ytLimfix)
x_a
x-a
x~a
7. Lim:"
--+ 00
X~O
8. Lim..£.n
x-co x
=0
From theorems 2 to 6 we can deduce that it does not make any
difference whether we:
(i) first simplify and then take the limit (left-hand side of the
theorem); or
(ii) first take the limit and then simplify.
Examples
[9 has nothing to do with x]
I. Lim 9 = 9
x~2
Lim 9x
2.
x~2
= 9 Lim x
x~2
= 9(2)
= 18
;~~
3.
=
+ ~)
(3x
;~~ 3x + ;~~ ~
= 3(2)
+1
=8
4.
Lim (3x . 4x)
x~2
= Lim
12x
2
x~2
= 12(2)2
= 48
Lim 16x
x~2 4x
5.
2
Lim 16x2
x---+2
Lim4x
x~2
=
or
= Lim 4x
x~2
=8
16(2)2
4(2)
=8
2
4
6. L'1m -::2
x~O
x
---t 00
7. Lim .§.=O
X-+OO x
1.2 LIMITS OF THE INDETERMINATE FORMS
Q. QQ. 00 00' 0 x 00
0' 00'
,
1.2.1 Factorising or dividing limits of the form ~
Lim x
2
x~2
4
-
[form Qol
x - 2
. (x - 2)(x + 2)
L 1m
x~2
(x - 2)
Lim x
x~2
x
or x - 2
Ix
2
x2
+2
+ Ox
-
, +
+2
-
4
2x
2x - 4
2x
4
2+2
4
1.2.2 Dividing limits of the form
~
To eliminate division by 00 we must divide the numerator and
denominator by the highest power of x in the denominator. From this
we can deduce that this method can only be applied to algebraic
expressions. For example:
.
L1m
x~oo
Lim
x-co
X4
+..d2x 2 - 1
3x' - 6
[form ~l
I+!i-.l.
x
3 - i..
x4
x4
1+0-0
3-0
1
'3
3
It is possible to see the answers to this type of problem straight away. If
the highest power of x in the numerator and denominator is equal, the
answer wilI be the coefficients of the highest powers. For example:
Lim 3x~ + 2x - 1
x~oo
[form
+ 16
4x - x
00]
00
3
4
If the highest power of x is in the numerator, the answer will be
For example:
-+ 00.
· 3x4 - 6
LI m - - +2
x~oo X
00
-+
If the highest power of x is in the denominator, the answer will be O.
For example:
L'm 3x4
I
x~oo
-
x5
6x + 12
+ 4x3
o
1.3 L'HOSPITAL'S RULE
.
fix)
.
f'(x)
Lim
- () = Lim
g'()
x~a g X
x~a
X
(0)
.
fix)
.
f'(x)
(00)
-0 and x~oo
Lim g- X() = x~oo
Lim g'()
X
00
We can apply this rule if:
•
•
•
•
g'(x) =J 0
fix) and g(x) can be differentiated in the vicinity of a .
Limf(x) = 0 and Lim g(x) = 0 :.
x--a
§
X-+OO
Lim fix)
-+ 00
x-oo
and Lim g(x)
-+ 00 :. 00
x-oo
00
L'Hospital's rule can only be used if the quotient is of the form
§ or ~ have been changed.
must be changed to §or ~ .
The rule must be applied till the forms
Forms
00 -
00,
0 x
00
or
00
+
00
4
§or ~.
Examples
· x2 - 4
LI m - 1.
x~2 x- 2
[form
· 2x - 0
L Im~
§]
[ L'Hospital]
x~2
2(2)
-1-
4
2.
· 3x2 + 2x - I
L 1m
4x
2
- I
x-co
Lim 6x
x~oo
+2
[Form : ; the answer will be~]
[ L'Hospital]
8x
Lim Q
[ L'Hospital]
x~oo 8
6
8
3
4
3.
Lim x 2 In x
[form 0 x - 00]
X~O
Lim Inlx
X~O
:;2
!
Lim
-!z
[ L'Hospital]
X~O ~
· I
_x 3
LI m - x X~O x
2
2
Lim (-;)
X~O
o
4.
Lim
x~l
(2:.- In x
Lim x-I
x~1 In x
· I
L1m,
x-I
x
_1_)
In x
[form 00 - 00]
[ L'Hospital]
Lim 1 x .!
x~l
I
5
5.
2
· f!'
LIm
-
x~oo X
x
+x
[form
Lim f!'
+ x2
[00]
00
X
x~oo
00
+ 00]
+ 2x
Lim f!'
1
x-oo
---> 00
6.
·
x2
LIm eX - 1
[8]
L'Im2x
[L'Hospital]
x~O
x-+O eX
0
T
0
· In (e 3x - 3x)
LIm
x~O
x
3x - 3]
_1_[3e
L' e3x - 3x
1m
1
7.
[8]
[ L'Hospital]
x~O
3e 3x - 3
e - x
·
Llm~
x-o
0
EXERCISE 1.1
Determine the value of each of the following limits:
2
1. Lim x
x~2
3. Lim
x-
+x
x - 2
2x3
2x
-~
5. L'1m x
3
- 6
+ 3x2
+3
x
-
2
-
9x + 9
x-I
x-I
2
x
7. Lim~
3
x 8
- 2. L'I .m
x~2 X - 2
x- 4
4. L"Im--x~4
3
2
x +1
6. Lim X
l
x~oo
8. L'1m
x~oo
x---+oo X
9. Lim sin 4x
x~o
x
..;x -
+1
X +x
1
~
2
10. Lim tanX x
x~o
6
11. Lim ~in x
X~O Sin 2x
12. Lim cos ~ - I
X~O
x Sin X
13. Lim x
14. Lim x
X~O
?OS ax
Sin ax
X~O
15. Lim ax + b
x~oo ex + d
17. Lim
X~O
X~~
x2
18. Limx~oo In x
--l.!LL
cosec x
20. Lim xe-x
X~O
X~O
+I
16. Lim (sec x - tan x)
19. Lim x 2 In sin x
21. L"1m
t sin x
X
X~O
ex .-
. e2x - I
22. Llm-.X~O
Sin x
e- X
Sin X
. In (e 2x - 2x)
23. Lim
X~O
2x
24.
. x+2
25. L"1m 2 cos
Sin X
26. Lim
L' (x
1m
h~O
X~O
x~"
+ h)2
-
x2
h
(...L
- ..,L)
2x
Sin X
Answers
1. 5
2. 12
9
3. 4
4. 4
5. -8
6.
7. 0
8. 0
9. 4
--> 00
I
10.
11. 2"
12.
13. 1
a
14. 0
15. f!e
16. 0
17. 0
18.
19. 0
20. 0
21. 2
22. 2
23. 0
24. 2x
25. 0
26.
--> 00
I
2"
--> 00
1.4 CONTINUITY
Continuous means without interruption. If there is an interruption at a
certain point in a graph, the function or relation becomes discontinuous at that point.
A function I is defined as continuous at a if Lim I(x) = I(a).
x~a
Therefore I is continuous at a if:
• I
is defined at a.
7
•
Limf(x) exists,
•
Limf(x)=f(a).
x~a
x~a
Example 1
Determine whether fwithf(x)
= ~ ~i
is continuous at x
= 2.
Solution
f(x) =
f(2)
=
~ ~ 24
22 - 4
2- 2
0
= (5
Thus fix) is not defined at 2 and f is discontinuous at 2.
It is not necessary to calculate the limit,
' x2 - 4
L1 m - x~2 X - 2
'
= L1m
x---+2
.ex----zJ
ex + 2)
tv-A-'r
~
...)
=2+2
=4
Graphically we can represent it as follows:
y-axis
4
2
x-axis
2
FIG. 1.1
8
Example 2
Determine whether fwithf(x) = 3x 2
-
4 is continuous at x = 2,
Solution
f(x) = 3x 2
f(2)
=
- 4
3(2)2 - 4
=8
Lim (3x 2
-
4)
x..... 2
= 3(2)2 - 4
=8
.., f is continuous at x = 2,
Example 3
x2 -
Determine whether fwithf(x) =
{
8
¥
16 'f
x::-T I
4
X
if x = 4
is continuous at x = 4,
Solution
f(4) = 8
' m
x2 - 16
-LI
[0]
-
4
0
' (x - 4)(x + 4)
L 1m
x ..... 4
(x - 4)
x ..... 4
X -
4+4
8
... f
is continuous at x = 4,
EXERCISE 1.2
Determine whether each function is continuous at the point with xvalue given in brackets:
1. f(x)
= 3xZ - 5
2. f(x) = ~
(2)
(0)
9
3. f(x) =
~
(1)
4. f(x)
= _\_x_
3:x + I
5. f(x)
= x2 + x -
6. f(x)
= x - 27
X2=9
(3)
J -=- \
(2)
(0)
(x _ 1)2
3
7. f(x) =
8. f(x)
=
9. f(x) =
10. f(x)
\X_(x + hr - x
2
2
=4_
11. f(x) =
12. f(x)
(1)
2
(2)
2
(0)
4 - x2
Vx2 + 12
{X +
3 if X =1= - 3
ifx=-3
o
= {~.;
i
o
13. f(x) = {024--
(2)
V;
if X
=1= -
(-3)
2
if X = - 2
(- 2)
if X =1= + 4
if X = 4
(4)
Answers
Continuous: Numbers 1; 3; 4; 7; 9; 11; 13
Discontinuous: Numbers 2; 5; 6; 8; 10; 12
10
Module 2
Differentiation
Objectives and overview
On completion of this section, you should be able to determine the
differential coefficients (or derivatives) of:
I. algebraic functions of the form f(x) = axn where n E R;
2. algebraic functions of the formf(x) = ~ ~ ~; where a, b, c and
d are constant; and
3. sin x and cos x.
2.1 DIFFERENTIATION FROM FIRST PRINCIPLES
2.1.1 Algebraic functions
The formula to calculate the gradient of a tangent to a curve or the
differential coefficient of a function namely Lim f(x + h) - f(x) was
deduced in N4.
h~O
h
The binomial theorem, namely
n
n
n - Ib
n(n - 1) n
(a + b) = a + na
+ 2! a
an -
3b
3
-
2b2
+
n(n - 1)(n - 2)
3!
+ ..., is also very handy.
Remember: 2! = 2 x 1 and 3! = 3 x 2 x 1 and 5! = 5 x 4 x 3 x 2 x I
Examples
I.
and
y=axn
"0 f(x) = axn
f(x + h) = a(x + ht
= a[xn + nxn
= a~ +
(1)
-
naxn -
11
+ n n; x n - 2 h2 + ... J
1 h + n(n; 1) a~-2 h2 +
oj
1h
0
•
f(x + h) - f(x)
= axn + nax n - Ih + n(n -; 1) ax n -
and
= naxn - Ih
+ n(n -;
1) axn -
2h 2
2h 2
+ ... _ axn
+ ... _ nax
. . ¥ = Limf(x + h) - f(x)
h~O
X
h
· naxnL 1m
=
+ -2-ax
n(n - I) n- 2h2 + ...
Ih
h
h~O
= Lim #(naxn -
I
+ powers of h)
#
h~O
=naxn -
[§J
I
fix) = 3 !2~
2
2.
. f(
..
x
+
h)
2 +x +h
= 3 - 2(x + h)
2+x+h
=3-2x-2h
... f(x
+ h)
- fix)
_ 2+x+h
-3-2x-2h
2+x
3-2x
(2 + x + h)
(3 - 2x)
(2 + x)
(3 - 2x -2h)
= 3 - 2x - 2h x (3 - 2x) - (3 - 2x) x (3 - 2x - 2h)
6 - 4x
+ 3x -
2x2
+ 3h -
2hx - (6 - 4x - 4h
(3 - 2x)(3 - 2x - 2h)
7h
= (3 - 2x)(3 -2x - 2h)
f'ex) = Limf(x + h) - f(x)
h
h~O
= Lim
h~O
= Lim
h~O
7h
h(3 - 2x)(3 - 2x - 2h)
7
(3 - 2x)(3 - 2x - 2h)
7
= (3 - 2x)2
12
+ 3x -
2x2
-
2hx)
vX+2
fix) =
3.
I
= (x + 2)2
+ h)
... fix
I
= (x
+ h + 2)2
= [(x +
2)
+
I
h]2
1
1(_*)
= (X + 2)2 + l(X + 2)-'h + Y
I
I
1
= (X + 2)2 + ~(X +
d f()
dx x =
L·
1m
fix
+ h)
h
h~O
I
1
3
+ 2)-2h2 + ...
k(x
3
+ 2)-2 h2 +... -
(x
1
+ 2)2
h
h~O
. h[!(x
LIm
+ 2)-! -
k(x
+ 2)-~h + powers of h]
h
h~O
2(x
3
+ 2)-2h2 + ...
- fix)
= Lim (x + 2)2 + !(x + 2P h -
=
1
I
2)-2h - g(X
(x
I
+ 2)"2
2.1.2 sin x and cos x
Let y = sin x . . .
CD
Let bx be a small increment in the x-direction and by be an increment
in the y-direction.
+ by = sin (x + bx) ... @
@ - CD: by = sin (x + bx) - sin x
. oy _ sin (x + ox) - sin x
f3\
... y
. . ox -
ox
. . . \:!.J
! B sin A 2" B
sin x = 2 cos ex + o{ + X) sin ex + o{ - X)
But sin A - sin B = 2 cos A
sin (x
+ bx)
-
= 2 cos
e{ + o{) sin o{
13
CD:
Substitute into
oy
ox =
=
2 cos (x
+ ,) sin ¥
ox
2 cos (x
= cos
(x
o
+ {) x
{j
+ {)
x
sin ~
T
sin~
~2
2
. Ql
. [
ox
... Lim
~ = Lim cos(x + 2) x
ox-O uX
ox-O
dy
... -d = cos (x
x
0
+ -2)
sin
¥]
bX
T
. sin x
x 1 [because Lim - - = 1]
x-O
x
= cos x
Let y = cos x . . .
Then y
+ 8y
= cos (x
CD
+ 8x) ... Q)
+ 8x) - cos x
Q) -
CD:
8y
From
CD:
8y = -2 sin (x +
CD
= cos (x
...
But cos A - cos B = -2 sin A! B sin A 2' B
. .. i!:
ox = -
=
2 sin (x
-sin (x
o{) sin o{
~
sin ~
+ Q:!)
2 x ~
ox
~
+ U{)
x
sin ~
h
2
2
... Lim
ox_O
sin
¥o = - Lim [sin (x + 0{) + ~
]
~
X
T
ox-O
. dy
.
.. dx = -sm x
EXERCISE 2.1
Use first principles to find the differential coefficient of each of the
following:
= x l8
1.
y
3.
fix) =~
5.
y
= ax 2 + bx + c
2. j(x)
I
= X!
I
= (3x)'J
4.
y
6.
fix)
14
2
=x-
3
7. y = x + 2
8. y
x-I
9. f(x)
11. y =
13. y
3-x
= 2+x
= 2x+
1
2 - 3x
10. fix) =
f~~
v'X+l
12. f(x) =
vx 1+ 1
= sin
2x
14. y = 2 cos 4x
1
15. fix) = V16/2 + 4
Answers
10.
3.
'xY
1
(3x)3
7. (x -3
_ 1)2
4.
2
3x3
-2
6.
(x - 3)2
2 -2
1. 18x l7
5. 2ax
--2
8.
3
(I - x)2
11.
13. 2 cos 2x
+b
-5
(2
-2
7
+ x)2
1
2vX+T
14. -8 sin 4x
9. (2 _ 3X)2
12.
-1
3
2(x + 1)"2
15. 0
Objectives and overview
On completion of this section, you should be able to:
1. determine the derivatives of tan x, cot x, sec x and cosec x;
2. apply the chain rule in combination with the quotient and
product rules on functions containing combinations of
algebraic and trigonometric functions;
3. use logarithms to simplify expressions to make differentiation
easier or possible;
4. define and differentiate implicit functions;
5. differentiate the six inverse trigonometric functions;
6. sketch the six inverse trigometric functions; and
7. do successive differentiation.
2.2 DIFFERENTIATION TECHNIQUES
2.2.1 Standard differential coefficients (and rules)
1. fx x n = nxn
-
I
2. fx kx n = kfxxn
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d
4. dx x
3. fx k = 0 [k is a constant]
6. fx aX
7. .!L log x = .!Llnx = 1
dx
e
dx
x
d .
9 . dx SIn
X
= cos
=
I
= aX In a or aX logea
8. fx loglOx = 0,:34
10. fx cos x = -sin x
x
II. fx tan x = secz x
12. fx sec x = sec x tan x
13. fx cot x = -cosec z x
14. fx cosec x = -cosec x cot x
15. fx [/(x) ± g(x)] = fxj{x) ± fx g(x)
Example 1
Findf'(x) ifj{x) = 2ex + ~
+ 3 sin
x -In
XZ.
Solution
f(x) = 2ex
. f'()
..
x
x - In X Z
+ ~ + 3 sin
d
2 d -I + 3dx
d·
= 2 dx
e + dx x
SIn
X
X -
2dx
d I
nx
= 2ex + 2(-lx- z) + 3 cos x -2(~)
= 2e
X
2
- Xi
+ 3 cos x
2
- X
Example 2
Find
Zif
y = 2 log3x
+ 2x
-
J3X + ~~~ ~ .
Solution
y = 2 log3x
+ 2x
. .!L _.!L [ 2InIn3x
. . dx Y - dx
= I; 3
+
fx In x
-
2x
J3X + ~~~ ~
1"13 1
]
- v.)
X + tan x
+Y
X
= -L
In 3' 1
x + 2 In
In 2 -
2-
v'3 fx x! + sec zx
v'3 . 12 x -1 + sec z x
v'3 + secZ x
= x In2 3 + 2xI2
n - 2JX
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Example 3
Calculate the x-value of the point at which the gradient of the tangent
to y = In x 3 is 2.
Solution
The gradient of the tangent is 2.
. dx
d In x 3 =
..
2
.". fx 3 In x
= 2
3.~
= 2
.".
.
3
.. x = 2"
EXERCISE 2.2
Find
I.
y
3.
y
5.
1x if:
= 2x2 + 3x x
x
= 2e + 3a x
4
+~
2 In x
y = 2e - 3 logex
4. y
= 3 log4x - 2 In ,jX
+ 2 log3x2
Differentiate with respect to x:
sin 2x
6. 2 cosx
7 _3
cos x
9. Calculate ~~ if s = ut + !at 2 .
2
;~ if y =
2_
. cot x
8. ax +;x + c
10. Calculate
v'I6X - ~
2. y =
m 2x 2
11. Calculate f'ex) if f(x)
+ 2mx
= yt24X -
12. Calculate f'(4) if f(x) = ,jX (1
13. Calculatef'(O) iff(O)
+ m2 _
,jX + 3ex .
- 3mx2
2 In
x2 .
+ xf
= (1 + tan 2 0) cos
O.
14. Calculate the gradient of the tangent to
y = sin x.+ cos x at x = 1!:4'
Sill X
IS. Calculate the x-values of the points at which the gradient of the
tangent to y = 2 cos x is 1 (0 ~ x ~ 27r).
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